Ionisation constant of each HA (weak acid) and BOH (weak base) are $3.0 x 10^{- 7}$ each at 298K. The percentage degree of hydrolysis of BA at the dilution of 10L is :

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Solution

The correct option is A 25
$\frac{h}{1 - h} = \sqrt{K H} = \sqrt{\frac{k w}{K_{a} k_{b}}} = \sqrt{\frac{10^{- 24}}{(3 \times 10^{- 7})^{2}}} = \frac{1}{3} \Rightarrow h = 0.25$

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Similar questions

Ionization constants of weak acid HA and weak base BOH are $3.0 \times 10^{- 7}$ each at 298 K. The percent degree of hydrolysis of their salt at the dilution of 10 L is :

Q. The percentage degree of hydrolysis of a salt of weak acid

(H A)

and weak base

(B O H)

in its

0.1 M

solution is found to be

10 %

. If the molarity of the solution is

0.05 M

, the percentage hydrolysis of the salt should be:

Q. What will be the

p H

and

% α

(degree of hydrolysis) respectively for the salt of weak acid and weak base

(B A)

0.1 M

concentration?
Given :

K_{a}

for

H A = 10^{- 6}

and

K_{b}

for

B O H = 10^{- 6}

Q. The dissociation constant of a weak acid HA is

4.9 \times 10^{- 7}

. Percentage ionisation :

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Ionisation constant of each HA (weak acid) and BOH (weak base) are 3.0x10−7 each at 298K. The percentage degree of hydrolysis of BA at the dilution of 10L is :

The correct option is A 25h1−h=√KH=√kwKakb=√10−24(3×10−7)2=13⇒h=0.25

Ionisation constant of each HA (weak acid) and BOH (weak base) are $3.0 x 10^{- 7}$ each at 298K. The percentage degree of hydrolysis of BA at the dilution of 10L is :

The correct option is A 25
$\frac{h}{1 - h} = \sqrt{K H} = \sqrt{\frac{k w}{K_{a} k_{b}}} = \sqrt{\frac{10^{- 24}}{(3 \times 10^{- 7})^{2}}} = \frac{1}{3} \Rightarrow h = 0.25$